A craft brewer wants to use an old mechanical spring balance to weigh bags of specialty hops. The scale dial is marked in kilograms (kg\text{kg}kg) instead of Newtons (N\text{N}N).
The scale has a zero-calibration error: with no load hanging from it, the scale displays a reading of 0.8 kg0.8\text{ kg}0.8 kg.
Calculate the weight (in Newtons) corresponding to this zero-error reading of 0.8 kg0.8\text{ kg}0.8 kg on Earth, where the gravitational field strength is g=9.8 N/kgg = 9.8\text{ N/kg}g=9.8 N/kg.
To verify the scale, the brewer hangs identical sealed cans of malt extract from the scale, placing them inside a lightweight canvas pouch of negligible mass. Each can is labeled as having a mass of 0.4 kg0.4\text{ kg}0.4 kg.
The readings are recorded in the table below:
| Number of cans | Scale reading (kg) |
|---|---|
| 0 | 0.8 |
| 1 | 1.2 |
| 2 | 1.6 |
| 3 | 2.5 |
| 4 | 2.4 |
| 5 | 2.8 |
| 6 | 3.2 |
Identify the anomalous data point in the table. Briefly explain why it is anomalous and how you would deal with it when plotting a line of best fit.
State the relationship between the number of cans and the scale reading (excluding the anomalous point).
The brewer recalls that six cans gave a scale reading of 3.2 kg3.2\text{ kg}3.2 kg. He calculates:
mass of six cans=6×0.4 kg=2.4 kg \text{mass of six cans} = 6 \times 0.4\text{ kg} = 2.4\text{ kg} mass of six cans=6×0.4 kg=2.4 kgand notes that:
3.2 kg−0.8 kg=2.4 kg 3.2\text{ kg} - 0.8\text{ kg} = 2.4\text{ kg} 3.2 kg−0.8 kg=2.4 kgHe concludes: "I can use this scale as normal! All I need to do is subtract 0.8 kg0.8\text{ kg}0.8 kg from each reading to get the correct mass."
He hangs a heavy sack of hops from the scale. The scale reading is 6.8 kg6.8\text{ kg}6.8 kg. He concludes that his sack must have a mass of exactly 6.0 kg6.0\text{ kg}6.0 kg.
Suggest three reasons why the brewer's conclusions might be incorrect.
